Answer
$4512~$ liters of oil leaks out during the first hour.
Work Step by Step
We can find the amount of oil that leaks out during the first 60 minutes:
$\int_{0}^{60}r(t)~dt = \int_{0}^{60}100e^{-0.01~t}~dt$
Let $u = -0.01~t$
$\frac{du}{dt} = -0.01$
$dt = -\frac{du}{0.01}$
When $t = 0$, then $u = 0$
When $t= 60$, then $u = -0.6$
$\int_{0}^{-0.6}(100e^u)~(-\frac{du}{0.01})$
$=~-\int_{0}^{-0.6}10,000~e^u~du$
$=~\int_{-0.6}^{0}10,000~e^u~du$
$=10,000~e^u~\vert_{-0.6}^{0}$
$=10,000~(e^0-e^{-0.6})$
$=10,000~(1-\frac{1}{e^{0.6}})$
$= 10,000~(0.4512)$
$4512~$ liters of oil leaks out during the first hour.
